Let the function space $A$ denote all functions $f : [0, 1) \to [0, 1)$ such that, for some set $Z$ of Lebesgue measure zero, the derivative $f'$ exists on $[0, 1) \setminus Z$ and $|f'| = 1$ there.
Let the function space $B$ denote all functions $f : [0, 1) \to [0, 1)$ such that for all $a < b$ in $[0, 1)$, the portion $G_f(a,b) = \{(x,f(x)) ∈ \mathbb R^2 \mid a \le x < b\}$ of the graph of $f$ that lies above $[a, b)$ has Hausdorff $1$-measure $H_1(G_f(a,b))$ equal to $\sqrt 2(b - a)$.
(Note that we do not assume functions to be continuous.)
Does $A = B$?
Edit: The following is false, as was shown to me by Christian Remling by an easy application of the Cantor function: "It is clear to me that $A$ is a subset of $B$." (Wrong!)
- So the real question is whether the reverse inclusion holds.
Let $F$ be a fat Cantor set, and let
$$ f(x) = x + \mathbf{1}_F(x). $$
Then for any $[a, b) \subseteq [0, 1)$,
\begin{align*} G_f([a,b)) &= \{ (x, x) : x \in [a, b) \setminus F\} \cup \{(x, x+1) : x \in [a, b) \cap F \} \end{align*}
and hence we get
\begin{align*} H_1(G_f([a,b))) &= \sqrt{2}\operatorname{Leb}([a, b) \setminus F) + \sqrt{2}\operatorname{Leb}([a, b) \cap F) \\ &= \sqrt{2}\operatorname{Leb}([a, b)) \\ &= \sqrt{2}(b - a). \end{align*}
This shows that $f$ is a member of $B$.
On the other hand, $f$ is not differentiable at each point in $F$. Therefore, $f$ does not lie in $A$.